If $\xi$ is a root of multiplicity $m$ of $f(x)=0$, then show that $\xi$ is a root of multiplicity $2m-1$ of the equation $$h(x)=f(x+f(x))-f(x)=0$$
And the conclude that $\xi$ is a simple root of the equation $$r(x)=\frac{f^2(x)}{h(x)}=0$$
Discuss about the disadvantages of applying Newton–Raphson method to $r(x)$.
If $\xi$ is a root with multiplicity $m$ of $f(x)$, then $f(x)=(x−\xi)^m\vartheta(x)$, with $\vartheta(\xi)≠0$. So here $$h(x)=f(x+f(x))-f(x)=$$
$$(x+(x−\xi)^m\vartheta(x)−\xi)^m\vartheta(x+(x−\xi)^m\vartheta(x))-(x−\xi)^m\vartheta(x)=0$$
I don't know how to conclude that $\xi$ is a root of multiplicity $2m-1$ of the equation $h(x)=0$ and so cannot use that to show that it's a simple root of $r(x)$.
Use the Taylor polynomial with remainder term of $h$, $$ f(x+v)-f(x)=f'(x)v+O(v^2)\implies h(x)=f'(x)f(x)+O(f(x)^2) $$ which has the required multiplicity in its first term, $m-1$ in $f'$ at $ξ$ and $m$ in $f$, and also a larger multiplicity in the error term.